Proposed Problem

Click the figure below to see the complete problem 413 about Cyclic Quadrilateral, Orthocenter of a triangle, Parallelogram, Concurrency, Congruence, Altitude.

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Complete Problem 413

Level: High School, SAT Prep, College geometry

## Monday, January 4, 2010

### Problem 413: Cyclic Quadrilateral, Orthocenter, Parallelogram, Concurrency, Congruence

Labels:
altitude,
concurrent,
congruence,
cyclic quadrilateral,
orthocenter,
parallelogram,
triangle

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Can you show that, in general, two quadrilaterals HFEG and ABCD have the same area?

ReplyDelete1. BCFE is a parallelogram

ReplyDeleteBE and CF cut circle (ABCD) at E’ and F’ .

Since E is the Orthocenter of triangle ABD , so E’ is the symmetric point of E over AD

Similarly F’ is the symmetric point of F over AD

BE//CF

BCE’F’ and BE’F’C are isosceles trapezoids and angle(BCF)=angle(CF’E’)=angle(EFF’)

So BC//EF and BCFE is a parallelogram

AGHD is a parallelogram

We have AG // HD ( both line perpen. to BC )

AG and HD cut the circle at G’ and H’ .

With the same logic as above we have GG’H’H and DH’G’A are isosceles trapezoids and GH //AD

So AGHD is a parallelogram

In the same way ABHF and CDEG are the parallelograms

2. Diagonals of a parallelogram bisect each other at mid point . Appling this propertie in above 4 parallelograms we will get the result.

3. Opposite sides of a parallelogram are congruence and opposite angles of a parallelogram are congruence.

( properties of parallelogram) . Applying these properties in above 4 parallelograms we have ABCD and HFEG have 4 sides and 4 internal angles congruence to each other . So they are congruence

Peter Tran

Problem 413

ReplyDeleteSince the problem 408 follows that BC=//EF , GE=//CD, GH=//AD,AB=//HF so BCFE, BHFA, AGHD ,ABHF are parallelograms.The (AH,BF),(GD,EC),(BF,EC),(BF,AH) intersecting at their mid. Which is the same point.

Triangle ABC=triangle HFE and triangle ACD=triangle HEG so quadr.ABCD=quadr.EFHG.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE